Projectiles: launch angle, launch height and clearing a wall Problem set A One question at a time

A pebble is thrown horizontally at 8 m s⁻¹ from a height of 20 m above level ground.

1. (a) Find the time it takes to reach the ground.

   • Vertical position \(y = 20 − 4.9t^{2}\) (it starts with no vertical speed).
   • Solve \(20 − 4.9t^{2} = 0\): \(t = 2.02\) s.

2. (b) Find the horizontal distance from the launch point to where it lands.

   • Horizontal position \(x = 8t\).
   • \(x = 8 \times 2.020 = 16.2\) m.

A cricket ball is projected from the origin with speed 16 m s⁻¹ at 50° above the horizontal. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up.

1. (a) Write down the horizontal and the vertical component of the initial velocity.

   • \(16\cos 50^\circ = 10.3\), \(16\sin 50^\circ = 12.3\).

2. (b) Write down the velocity vector at time \(t\) seconds.

   • Horizontal: constant. Vertical: decreases by 9.8 every second.
   • \(\mathbf{v} = \begin{pmatrix} 10.28 \\ 12.26 − 9.8t \end{pmatrix}\).

3. (c) Write down the position vector at time \(t\) seconds.

   • Integrate, with zero constants because it starts at the origin.
   • \(\mathbf{r} = \begin{pmatrix} 10.28t \\ 12.26t − 4.9t^{2} \end{pmatrix}\).

A rugby ball is projected from level ground with speed 24 m s⁻¹ at 25° above the horizontal.

1. (a) Find the time it takes to reach its highest point.

   • Vertical component \(24\sin 25^\circ = 10.14\).
   • \(10.14 − 9.8t = 0\), \(t = 1.03\) s.

2. (b) Find its greatest height.

   • \(y = 10.14(1.035) − 4.9(1.035)^{2} = 5.25\) m.

3. (c) Find the horizontal distance from the launch point to where it lands.

   • Landing time \(2 \times 1.035 = 2.070\) s.
   • Horizontal component \(24\cos 25^\circ = 21.75\).
   • \(x = 21.75 \times 2.070 = 45.0\) m.

An arrow is projected from a point 1.6 m above level ground with speed 30 m s⁻¹ at 20° above the horizontal.

1. (a) Find the time it takes to reach the ground.

   • Vertical position \(y = 1.6 + 10.26t − 4.9t^{2}\).
   • Polynomial solver: \(y = 0\), \(t = 2.24\) s (the positive solution).

2. (b) Find the horizontal distance from the launch point to where it lands.

   • Horizontal component \(28.19\).
   • \(x = 28.19 \times 2.240 = 63.1\) m.

3. (c) Find its greatest height above the ground.

   • Top at \(10.26 − 9.8t = 0\), \(t = 1.047\) s.
   • \(y = 1.6 + 10.26(1.047) − 4.9(1.047)^{2} = 6.97\) m.

A stone is thrown from a bridge. It is thrown horizontally at 6 m s⁻¹ and hits the water 2.5 seconds later. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up.

1. (a) Find its velocity as a column vector when it hits the water.

   • \(\mathbf{v} = \begin{pmatrix} 6 \\ −9.8t \end{pmatrix}\), so \(\mathbf{v}(2.5) = \begin{pmatrix} 6 \\ −24.5 \end{pmatrix}\).

2. (b) Find its speed when it hits the water.

   • \(|\mathbf{v}| = \sqrt{6^{2} + 24.5^{2}} = 25.2\) m s⁻¹.

3. (c) Find the angle below the horizontal at which it hits the water.

   • \(\tan\theta = \dfrac{24.5}{6}\), \(\theta = 76.2\)° below the horizontal.

A ball is kicked from level ground with speed 15 m s⁻¹ at 45° above the horizontal, towards a wall 4.5 m high that is 20 m away along the ground. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin at the launch point.

1. (a) Find the time at which the ball is above the wall.

   • Horizontal component \(10.61\).
   • \(10.61t = 20\), \(t = 1.886\) s.

2. (b) Find the height of the ball at that time.

   • \(y = 10.61(1.886) − 4.9(1.886)^{2} = 2.58\) m.

3. (c) Does the ball clear the wall? Give the difference between its height and the top of the wall.

   • \(2.58 < 4.5\).
   • The ball is 1.92 m below the top: it hits the wall.

A diver leaves horizontally at 3 m s⁻¹ from a point 12 m above level water.

1. (a) Find the time it takes to reach the water.

   • \(12 − 4.9t^{2} = 0\), \(t = 1.56\) s.

2. (b) Find how far from the point directly below the launch it lands.

   • \(x = 3 \times 1.565 = 4.69\) m.

3. (c) Find its speed and the angle below the horizontal as it reaches the water.

   • \(\mathbf{v} = \begin{pmatrix} 3 \\ −9.8t \end{pmatrix}\), \(\mathbf{v}(1.565) = \begin{pmatrix} 3 \\ −15.3 \end{pmatrix}\).
   • \(|\mathbf{v}| = 15.6\) m s⁻¹, \(\tan\theta = \dfrac{15.3}{3}\), \(\theta = 78.9\)°.

A javelin is released from a point 1.8 m above level ground with speed 28 m s⁻¹ at 35° above the horizontal. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin on the ground below the release point.

1. (a) Write down the position vector at time \(t\) seconds.

   • \(28\cos 35^\circ = 22.94\), \(28\sin 35^\circ = 16.06\).
   • \(\mathbf{r} = \begin{pmatrix} 22.94t \\ 1.8 + 16.06t − 4.9t^{2} \end{pmatrix}\).

2. (b) Find its greatest height above the ground.

   • \(16.06 − 9.8t = 0\), \(t = 1.639\) s.
   • \(y = 1.8 + 16.06(1.639) − 4.9(1.639)^{2} = 15.0\) m.

3. (c) Find the time it takes to land and the horizontal distance of the landing point from the origin.

   • Polynomial solver: \(y = 0\), \(t = 3.39\) s.
   • \(x = 22.94 \times 3.386 = 77.7\) m.

In a free kick the defenders form a wall. The ball is kicked from level ground with speed 20 m s⁻¹ at 25° above the horizontal. The wall of defenders is 1.9 m high and stands 9.15 m from the kick. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin at the kick.

1. (a) Find the height of the ball when it is 9.15 m from the kick, and say whether it clears the 1.9 m wall.

   • Time at 9.15 m: \(18.13t = 9.15\), \(t = 0.5048\) s.
   • \(y = 8.452(0.5048) − 4.9(0.5048)^{2} = 3.02\) m, which is above 1.9 m: it clears the wall.

2. (b) Find the speed of the ball at that moment.

   • \(\mathbf{v} = \begin{pmatrix} 18.13 \\ 8.452 − 9.8t \end{pmatrix}\), \(\mathbf{v}(0.5048) = \begin{pmatrix} 18.13 \\ 3.51 \end{pmatrix}\).
   • \(|\mathbf{v}| = 18.5\) m s⁻¹.

3. (c) Find how far from the kick the ball lands.

   • Landing time \(2 \times 0.8625 = 1.725\) s.
   • \(x = 18.13 \times 1.725 = 31.3\) m.

A tennis ball is projected from level ground with speed 20 m s⁻¹ at 50° above the horizontal.

1. (a) Find the two times at which the ball is 6 m above the ground.

   • \(y = 15.32t − 4.9t^{2}\).
   • Polynomial solver: \(15.32t − 4.9t^{2} = 6\), \(t = 0.459\) or \(t = 2.67\). One time is on the way up, one on the way down.

2. (b) Find the horizontal distance from the launch point at each of these times.

   • \(x = 12.86t\): \(12.86 \times 0.4590 = 5.90\) m and \(12.86 \times 2.668 = 34.3\) m.

3. (c) For how long is the ball more than 6 m above the ground?

   • It is above 6 m between the two times: \(2.668 − 0.4590 = 2.21\) s.

A ball is kicked from level ground at an angle of 45° above the horizontal, with speed \(V\) m s⁻¹, towards a wall 4 m high that is 20 m away along the ground. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin at the kick.

1. (a) Show that the height of the ball when it is 20 m from the kick is \(20 − \dfrac{3920}{V^{2}}\) metres.

   • \(\mathbf{r} = \begin{pmatrix} \dfrac{V}{\sqrt{2}}t \\ \dfrac{V}{\sqrt{2}}t − 4.9t^{2} \end{pmatrix}\).
   • \(\dfrac{V}{\sqrt{2}}t = 20\), so \(t = \dfrac{20\sqrt{2}}{V}\).
   • \(y = 20 − 4.9\left(\dfrac{20\sqrt{2}}{V}\right)^{2} = 20 − \dfrac{3920}{V^{2}}\).

2. (b) Find the least value of \(V\) for which the ball clears the wall.

   • The ball just clears the wall when \(20 − \dfrac{3920}{V^{2}} = 4\).
   • Solve on the calculator: \(V = 15.7\) m s⁻¹ (positive solution).
   • A larger \(V\) gives a greater height, so this is the least value.